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1D systems Solve the TISE for various 1D potentials • Free particle • Infinite square well • Finite square well • Particle flux • Potential step Transmission and reflection coefficients • The barrier potential Quantum tunnelling Examples of tunnelling • The harmonic oscillator 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 1 A Free Particle Free particle: no forces so potential energy independent of position (take as zero) d 2 E 2 2m dx Time-independent Schrödinger equation: 2 Linear ODE with constant coefficients so try exp( x) General solution: Combine with time dependence to get full wave function: 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 2 Notes • Plane wave is a solution (just as well, since our plausibility argument for the Schrödinger equation was based on this assumption). ( x) e i kx t • Note signs in exponentials: – Sign of time term (-iωt) is fixed by sign adopted in time-dependent Schrödinger Equation – Sign of position term (±ikx) depends on propagation direction of wave. +ikx propagates towards +∞ while -ikx propagates towards –∞ • There is no restriction on k and hence on the allowed energies. The states form a continuum. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 3 Particle in a constant potential General solutions we will use over and over again Time-independent Schrödinger equation: Case 1: E > V (includes free particle with V = 0 and K = k) d 2 V E 2 2m dx 2 2m E V d 2 2 2 K 0, K 2 2 dx Solution: Case 2: E < V (classically particle can not be here) 2m V E d 2 2 2 q 0, q 2 2 dx Solution: 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 4 Infinite Square Well Consider a particle confined to a finite length –a<x<a by an infinitely high potential barrier V V V(x) V 0 No solution in barrier region (particle would have infinite potential energy). In the well V = 0 so equation is the same as before x d E 2m dx 2 2 2 -a a General solution: Boundary conditions: Continuity of ψ at x = a: Continuity of ψ at x = -a: 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT Note discontinuity in dψ/dx allowable, since potential is infinite 5 Infinite Square Well (2) Add and subtract these conditions: Even solution: ψ(x) = ψ(-x) Odd solution: ψ(x) = -ψ(-x) Energy We have discrete states labelled by an integer quantum number 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 6 Infinite Square Well (3) Normalization Normalize the solutions Calculate the normalization integral N ( x, t ) dx 2 Normalized solutions are 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 7 Infinite Square Well (4) Sketch solutions Wavefunctions ( x ) Probability density ( x) 2 3 1 Note: discontinuity of gradient of ψ at edge of well. OK because potential is infinite there. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 8 Infinite Square Well (5) Relation to classical probability distribution Classically particle is equally likely to be anywhere in the box Quantum probability distribution is 1 n n x cos 2 x , n 1,3,5 a 2a 2 1 n n x sin 2 x , n 2, 4, 6 2 a But 1 Pcl ( x) 2a 2a cos 2 sin 2 1/ 2 so the high energy quantum states are consistent with the classical result when we can’t resolve the rapid oscillations. This is an example of the CORRESPONDENCE PRINCIPLE. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 9 Infinite Square Well (5) – notes • Energy can only have discrete values: there is no continuum of states anymore. The energy is said to be quantized. This is characteristic of boundstate problems in quantum mechanics, where a particle is localized in a finite region of space. • The discrete energy states are associated with an integer quantum number. • Energy of the lowest state (ground state) comes close to bounds set by the Uncertainty Principle: • The stationary state wavefunctions are even or odd under reflection. This is generally true for potentials that are even under reflection. Even solutions are said to have even parity, and odd solutions have odd parity. • Recover classical probability distribution at high energy by spatial averaging. • Warning! Different books differ on definition of well. E.g. – B&M: well extends from x = -a/2 to x = +a/2. Our results can be adapted to this case easily (replace a with a/2). – May also have asymmetric well from x = 0 to x = a. Again can adapt our results here using appropriate transformations. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 10 Finite Square Well Now make the potential well more realistic by making the barriers a finite height V0 V(x) I II III V0 Assume 0 E V0 x i.e. particle is bound Region I: 2017/5/8 -a Region II: Chang-Kui Duan, Institute of Modern Physics, CUPT a Region III: 11 Finite Square Well (2) Boundary conditions: match value and derivative of wavefunction at region boundaries: x a x a Match ψ: Match dψ/dx: Now have five unknowns (including energy) and five equations (including normalization condition) Solve: 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 12 Finite Square Well (3) Even solutions when Odd solutions when q k tan ka q k cot ka We have changed the notation 2017/5/8 Cannot be solved algebraically. Solve graphically or on computer into q Chang-Kui Duan, Institute of Modern Physics, CUPT 13 Finite Square Well (4) Graphical solution k0 = 4 a=1 Even solutions at intersections of blue and red curves (always at least one) Odd solutions at intersections of blue and green curves 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 14 Finite Square Well (5) Sketch solutions Wavefunctions ( x ) Probability density ( x) 2 Note: exponential decay of solutions outside well 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 15 Finite Square Well (6): Notes •Tunnelling of particle into “forbidden” region where V0 > E (particle cannot exist here classically). •Amount of tunnelling depends exponentially on V0 – E. •Number of bound states depends on depth of well, but there is always at least one (even) state •Potential is even, so wavefunctions must be even or odd •Limit as V0→∞: n k0 Solutions at ka 2 We recover the infinite well solutions as we should. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 16 Example: the quantum well Quantum well is a “sandwich” made of two different semiconductors in which the energy of the electrons is different, and whose atomic spacings are so similar that they can be grown together without an appreciable density of defects: Material A (e.g. AlGaAs) Material B (e.g. GaAs) Electron energy Position Now used in many electronic devices (some transistors, diodes, solid-state lasers) 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 17 Summary of Infinite and Finite Wells Infinite well Even parity Odd parity Finite well Infinitely many solutions 1 n cos x , n 1,3,5 a 2a 1 n n x sin x , n 2, 4, 6 a 2a n x Finite number of solutions At least one solution (even parity) Evanescent wave outside well. Even parity solutions Odd parity solutions q k tan ka q k cot ka 2017/5/8 n 2 2 2 En 8ma 2 Chang-Kui Duan, Institute of Modern Physics, CUPT q2 2m 2 k2 V0 E 2mE 2 18 Particle Flux In order to analyse problems involving scattering of free particles, need to understand normalization of free-particle plane-wave solutions. A exp[i(kx t )] dx 2 A dx 2 Conclude that if we try to normalize so that dx 1 2 we get A = 0. This problem is related to Uncertainty Principle: Momentum is completely defined Solutions: 2017/5/8 Position completely undefined; single particle can be anywhere from -∞ to ∞, so probability of finding it in any finite region is zero Normalize in a finite box Use wavepackets (later) Use a flux interpretation Chang-Kui Duan, Institute of Modern Physics, CUPT 19 Particle Flux (2) More generally: what is the rate of change of probability that a particle is in some region (say, between x=a and x=b)? b b * d Prab d *dx * dx dt dt a t t a Use time-dependent Schrödinger equation: 2017/5/8 a b x 2 2 i V ( x, t ) 2 t 2m x Chang-Kui Duan, Institute of Modern Physics, CUPT 20 Particle Flux (3) d Prab dt Interpretation: Particle flux at position x i * * j ( x, t ) 2m x x 2017/5/8 Flux entering at x=a minus a Flux leaving at x=b b x Note: a wavefunction that is real carries no current Chang-Kui Duan, Institute of Modern Physics, CUPT 21 Particle Flux (4) i * * j ( x, t ) 2m x x Check: apply to free-particle plane wave. Makes sense: # particles passing x per unit time = # particles per unit length × velocity So plane wave wavefunction describes a “beam” of particles. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 22 Particle Flux (5): Notes • Particle flux is nonlinear i * * j ( x, t ) 2m x x • Time-independent case: replace x, t with x • 3D case, x • Can use this argument to prove CONSERVATION OF PROBABILITY. Put a = -∞, b = ∞, then Prab N and d Prab dN j (a, t ) j (b, t ) j (, t ) j (, t ) dt dt If 0, then 2017/5/8 dN 0 dt Chang-Kui Duan, Institute of Modern Physics, CUPT 23 Potential Step V(x) Consider a potential which rises suddenly at x = 0: Case 1 V0 0 Case 1: E > V0 (above step) x < 0, V = 0 2017/5/8 x Boundary condition: particles only incident from left x > 0, V = V0 Chang-Kui Duan, Institute of Modern Physics, CUPT 24 Potential Step (2) Continuity of ψ at x = 0: Continuity of d at x 0 : dx Solve for reflection and transmission amplitudes: r 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT kK 2k , t kK kK 25 Potential Step (3) Transmission and Reflection Fluxes Calculate transmitted and reflected fluxes x<0 i * * j ( x) 2m x x x>0 1 ( x) eikx r e ikx 2 ( x) teiKx (cf classical case: no reflected flux) Check: conservation of particles 2017/5/8 k q 2 2 1 r t m m Chang-Kui Duan, Institute of Modern Physics, CUPT 26 Potential Step (4) Case 2: E < V0 (below step) Solution for x < 0 same as before V(x) V0 0 Solution for x > 0 is now evanescent wave Matching boundary conditions: Transmission and reflection amplitudes: Transmission and reflection fluxes: This time we have total reflected flux. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 27 Potential Step (5): Notes • Some tunnelling of particles into classically forbidden region even for energies below step height (case 2, E < V0). • Tunnelling depth depends on energy difference V0 E, 1 q 2m V0 E • But no transmitted particle flux, 100% reflection, like classical case. • Relection probability is not zero for E > V0 (case 1). Only tends to zero in high energy limit, E >> V (correspondence principle again). 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 28 Rectangular Potential Barrier V(x) Now consider a potential barrier of finite thickness: I II III V0 x 0 Assume 0 E V0 Region I: b Boundary condition: particles only incident from left Region II: u = exp(ikx) + B exp(−ikx) u = C exp(Kx) + D exp(−Kx) 2017/5/8 Region III: u = F exp(ikx) Chang-Kui Duan, Institute of Modern Physics, CUPT 29 Rectangular Barrier (2) Match value and derivative of wavefunction at boundaries: x b x0 Match ψ: 1+B=C+D C exp(Kb) + D exp(−Kb) = F exp(ikb) Match dψ/dx: 1 − B = K/(ik)(C − D) C exp(Kb) − D exp(−Kb) = ik/K F exp(ikb) Eliminate wavefunction in central region: 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 30 Rectangular Barrier (3) Transmission and reflection amplitudes: F For very thick or high barrier: |F|2 = Non-zero transmission (“tunnelling”) through classically forbidden barrier region. Exponentially sensitive to height and width of barrier. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 31 Examples of Tunnelling Tunnelling occurs in many situations in physics and astronomy: 1. Nuclear fusion (in stars and fusion reactors) V ( Ze)2 Barrier height ~ ~ MeV 4 0 rnucleus Repulsive Coulomb interaction thermal energies (~keV) at T 107 K Incident particles Assume a Boltzmann distribution for the KE, P E e E / kT Nuclear separation x Strong nuclear force (attractive) 2017/5/8 Probability of nuclei having MeV energy is e 1000 Fusion (and life) occurs because nuclei tunnel through the barrier Chang-Kui Duan, Institute of Modern Physics, CUPT 32 Examples of Tunnelling 2. Alpha-decay V Distance of α-particle from nucleus Initial α-particle energy α-particle must overcome Coulomb repulsion barrier. Tunnelling rate depends sensitively on barrier width and height. t 2 e2qa Explains enormous range of α-decay rates, e.g. 232Th, t1/2 = 1010 yrs, 218Th, t1/2 = 10-7s. Difference of 24 orders of magnitude comes from factor of 2 change in α-particle energy! 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 33 Examples of Tunnelling 3. Scanning tunnelling microscope V x A conducting probe with a very sharp tip is brought close to a metal. Electrons tunnel through the empty space to the tip. Tunnelling current is so sensitive to the metal/probe distance (barrier width) that even individual atoms can be mapped. Tunnelling current proportional to q 2 2m E V0 2 so and q 1 A t 2 a Vacuum Probe e2qa E V0 4eV 1 If a changes by 0.01A (~1/100th of the atomic size) then current changes by a factor of 0.98, i.e. a 2% change, which is detectable 2017/5/8 Material STM image of Iodine atoms on platinum. The yellow pocket is a missing Iodine atom Chang-Kui Duan, Institute of Modern Physics, CUPT 34 Summary of Flux and Tunnelling The particle flux density is i * * j ( x, t ) 2m x x Particles can tunnel through classically forbidden regions. Transmitted flux decreases exponentially with barrier height and width t 2 e2qa q 2 2m E V0 2 We get transmission and reflection at potential steps. There is reflection even when E > V0. Only recover classical limit for E >> V0 (correspondence principle) kK 2k , t kK kK k iq 2k , t E V0 r k iq k iq E V0 r 2 2 2 2 k2 K2 q E , E V0 ( E V0 ), V0 E ( E V0 ) 2m 2m 2m 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 35 Simple Harmonic Oscillator Example: particle on a spring, Hooke’s law restoring force with spring constant k: Mass m Force F kx k m Potential energy V ( x) 1 kx 2 1 m0 2 x 2 2 2 Angular frequency 0 = x V(x) Time-independent Schrödinger equation: Problem: still a linear differential equation but coefficients are not constant. Simplify: change to dimensionless variable 2017/5/8 m0 y 1/ 2 x, =2E/( 0 ) Chang-Kui Duan, Institute of Modern Physics, CUPT x 36 Simple Harmonic Oscillator (2) Asymptotic solution in the limit of very large y: ( y) exp( y 2 / 2) Try it: Suggests we substitute ( y) H ( y) exp( y 2 / 2) Equation for H(y): d 2H dH 2 y 1 H 0 2 dy dy 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 37 Simple Harmonic Oscillator (3) Solve this ODE by the power-series method (Frobenius method): H ( y) a p y p 0 p d 2H dH 2 y 1 H 0 2 dy dy Find that series for H(y) must terminate for a normalizable solution Can make this happen after n terms for either even or odd terms in series (but not both) by choosing Hence solutions are either even or odd functions (expected on parity considerations) 0 Label normalizable functions H by the values of n (the quantum number) 2017/5/8 Hn is known as the nth Hermite polynomial. Chang-Kui Duan, Institute of Modern Physics, CUPT 38 Simple Harmonic Oscillator (4) EXAMPLES OF HERMITE POLYNOMIALS AND SHO WAVEFUNCTIONS H0 y 1 0 y N 0e H1 y 2 y 1 y N1 2 ye y2 / 2 y2 / 2 2 y N2 4 y 2 e H2 y 4 y 2 2 2 y2 / 2 H 3 y 8 y 12 y 3 y N 3 8 y 12 y e 3 3 N0 , N1 , N 2 Hn y 2017/5/8 Nn y2 / 2 are normalization constants is a polynomial of degree yn Chang-Kui Duan, Institute of Modern Physics, CUPT 39 Simple Harmonic Oscillator (5) Wavefunctions High n state (n=30) wavefunction Probability density Pcl ( x) 1 a2 x2 • Decaying wavefunction tunnels into classically forbidden region 1 • Spatial average for high energy wavefunction gives classical result: 2 2 E m 0a another example of the CORRESPONDENCE PRINCIPLE 2 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 40 Summary of Harmonic Oscillator 1) The quantum SHO has discrete energy levels because of the normalization requirement 1 En n 0 , n 0,1, 2,3 2 2) There is ‘zero-point’ energy because of the uncertainty principle. E0 3) Eigenstates are Hermite polynomials times a Gaussian 0 2 ( y) H ( y) exp( y 2 / 2) 4) Eigenstates have definite parity because V(x) = V(-x). They can tunnel into the classically forbidden region. 5) For large n (high energy) the quantum probability distribution tends to the classical result. Example of the correspondence principle. 6) Applies to any SHO, eg: molecular vibrations, vibrations in a solid (phonons), electromagnetic field modes (photons), etc 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 41 Example of SHOs in Atomic Physics: Bose-Einstein Condensation 87Rb atoms are cooled to nanokelvin temperatures in a harmonic trap. de Broglie waves of atoms overlap and form a giant matter wave known as a BEC. All the atoms go into the ground state of the trap and there is only zero point energy (at T=0). This is a superfluid gas with macroscopic coherence and interference properties. Signature of BEC phase transition: The velocity distribution goes from classical Maxwell-Boltzmann form to the distribution of the quantum mechanical SHO ground state. 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 42 Example of SHOs: Molecular vibrations VIBRATIONAL SPECTRA OF MOLECULES Useful in chemical analysis and in astronomy (studies of atmospheres of cool stars and interstellar clouds). V(x) H2 molecule H H x Nuclear separation x SHO levels SHO very useful because any potential is approximately parabolic near a minimum 2017/5/8 Chang-Kui Duan, Institute of Modern Physics, CUPT 43